It is to show that $2,3, 1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]$ but not primes and that 1 and -1 are the only units.

Let $N$ be the norm map into $\mathbb{Z}$ and let u denote the unit, then because it is a homomorphism it follows that :

$N(xy)=N(x)N(y)$ and thus with $N(u)N(u^{-1})=N(uu^{-1})=N(1)$ it follows that the norm of u is a divider of 1 and thus also a unit of $\mathbb{Z}$. Therefore we have in $\mathbb{Z}[\sqrt{-5}]$: $\forall a,b \in \mathbb{Z}: N(a+b\sqrt{-5})=(a+b\sqrt{-5})(a-b\sqrt{-5})=a^{2}+5b^{2}=1 \ \ \Rightarrow u= \pm 1$

Now it is to show that $2,3,1-\sqrt{-5}, 1+\sqrt{-5}$ are irreducible over $\mathbb{Z}[\sqrt{-5}]: $

Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$ But since $1-\sqrt{-5}$ is not a quadratic remainder of $5$, there doesn't exist a solution for the equations $a^{2}+5b^{2}= \pm (1-\sqrt{-5})$ And the exactly same argument also works for $2,3$ and $1+\sqrt{-5}$.

Thus we have shown that $2,3,1\pm \sqrt{-5}$ are not reducible over $\mathbb{Z}[\sqrt{-5}]$.

Now we show that they are not prime:

Assume that 2 is a prime in $\mathbb{Z}[\sqrt{-5}]$, then because of $2\cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5})=6$ it must hold that $2|(1-\sqrt{-5})$ or $2|(1+\sqrt{-5})$. But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :

$(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.

Assume that 3 is a prime in $\mathbb{Z}[\sqrt{-5}$, then: $(1\pm \sqrt{-5}) = 3(a+b\sqrt{-5}) \Rightarrow 3b= \pm 1$ it follows that 3 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.

Assume that $1\pm \sqrt{-5}$ is a prime in $\mathbb{Z}[\sqrt{-5}]$, then: $3 = (1\pm \sqrt{-5})(a+b\sqrt{-5}) \Rightarrow (1\pm \sqrt{-5} )a = 3$ which is not solvable in $\mathbb{Z}$. And also $2=(1\pm \sqrt{-5})(a+b\sqrt{-5})=(1\pm \sqrt{-5})a$ which is also not solvable in $\mathbb{Z}$ and thus $(1\pm \sqrt{-5})$ can not be a prime in $\mathbb{Z}[\sqrt{-5}]$.

There are a number of errors. The norm is an integer. And the usual quadratic residues/non-residues are ordinary integers. For example to show $1+\sqrt{-5}$ is irreducible, calculate the norm. The only non-trivial factors are $2$ and $3$, and it is obvious we cannot have $x^2+5y^2=2$ (or $3$), no need to appeal to fancier stuff.
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André NicolasOct 29 '11 at 21:10

The proof that $1-\sqrt{-5}$ is not prime is not clear. What you should do is show say that $1-\sqrt{-5}$ divides $6$ (easy), but divides neither $2$ nor $3$. If it divided $2$, its norm $6$ would divide the norm of $2$, which is $4$, in the ordinary integer sense. But $6$ does not divide $4$. Neither does it divide $9$, the norm of $3$.
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André NicolasOct 29 '11 at 21:21

It could be clearer, but there are no mistakes. You should say something like this. Clearly $1$ and $-1$ are units. We show there are no others. If $u$ is a unit, then $uv=1$ for some $v$. Then $N(uv)=N(u)N(v)=1$ and therefore $N(u)=1$. But $N(a+b\sqrt{-5})=a^2+5b^2$, and this can be $1$ only for $a=\pm 1$, $b=0$.
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André NicolasOct 29 '11 at 21:35

@VVV: There is no need to "sign" your posts: every post you make has your user name on the bottom right corner, which acts like a "signature"
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Arturo MagidinOct 29 '11 at 22:26

1 Answer
1

The proof that $1-\sqrt{-5}$ is irreducible is incorrect. In addition to a minor (and alas, common) mistake in writing that makes what you write not what you intended to write, there is an assertion which is just plain false.

Explicilty, you write:

Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$

First: the use of $\Rightarrow$ is incorrect. What you have written is that there exist $a$ and $b$ in $\mathbb{Z}[\sqrt{-5}]$ for which the following statement holds:

If $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b) = \pm (1-\sqrt{-5})$.

What you actually wanted to write was that

If there exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b)=\pm(1-\sqrt{-5})$.

What's the difference? The first statement will be true if you can find an $a$ and a $b$ for which $N(1-\sqrt{-5})$ is not equal to $N(a)N(b)$! It will be true that the implication holds, because the antecedent will be false. So, exhibiting $a=7$ and $b=10578432$ makes the statement you wrote true. However, they are irrelevant towards the second statement (and towards establishing what you want to establish, namely, that no such $a$ and $b$ exist).

Second: this is incorrect. What you want to assume is that there exist $a$ and $b$ such that $1-\sqrt{-5} = ab$, and neither $a$ nor $b$ are units; you do not simply want to assume that the product of the norms of $a$ and $b$ equals the norm of $1-\sqrt{-5}$.

Third: Even so, you conclusion is nonsense. The norm of any element of $\mathbb{Z}[\sqrt{-5}]$ must be an integer. What you want to conclude is that $N(a)N(b) = N(1-\sqrt{-5}) = 6$, and then get a contradiction. The norm cannot equal $\pm(1-\sqrt{-5})$.

It's also nonsense to say "since $1-\sqrt{-5}$ is not a quadratic remainder modulo $5$". It's not even a remainder modulo $5$, because it's not an integer!

The argument about $2$ not being prime is likewise incorrect in its use of the norm, which quickly reduces to nonsense symbols being strewn around:

But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :
$(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.

This is nonsensical as written. I suspect you wanted to say something like

But there cannot exist $a,b\in\mathbb{Z}$ (not in $\mathbb{Z}[\sqrt{-5}]$) such that $2(a+b\sqrt{-5}) = 2a+2b\sqrt{-5} = 1-\sqrt{-5}$.

Same issues with the rest of the arguments. They are either terse to the point of nonsense, or contain incorrect or incoherent claims. I strongly urge you to write complete sentences, use words, and don't over rely on symbols. And to read your own arguments with a critical eye after you are done.